What is $\int 16 {sin}^{2} x {cos}^{2} x d x$ $int 16sin^2 xcos^2 x dx$ ?

Question

6647 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-10T12:46:48

$\int 16 {sin}^{2} x {cos}^{2} x d x$ $int 16sin^2 xcos^2 x dx$

$= 2 \int 2 \cdot {(2 sin x cos x)}^{2} d x$ $=2int 2*(2sin xcos x )^2dx$

$= 2 \int 2 \cdot {sin}^{2} 2 x d x$ $=2int 2*sin ^2 2xdx$

$= 2 \int (1 - cos 4 x) d x$ $=2int (1-cos 4x)dx$

$= 2 \int d x - 2 \int cos 4 x d x$ $=2int dx-2intcos 4xdx$

$= 2 x - 2 \cdot \frac{sin 4 x}{4} + c$ $=2x-2*(sin4x)/4+c$ , where c is integration constant

$= 2 x - \frac{1}{2} (sin 4 x) + c$ $=2x-1/2(sin4x)+c$

Answer 2 · 2018-04-10T12:47:00

$int \ 16sin^2x cos^2x \ dx = 2x - 1/2sin4x + C$

We want to evaluate the integral:

$I = int \ 16sin^2x cos^2x \ dx$

Using the identity:

$sin 2A-= 2sinAcosA$

We can write:

$I = int \ 4*4*(sinxcosx)^2 \ dx$
$\ \ = int \ 4(2sinxcosx)^2 \ dx$
$\ \ = int \ 4(sin2x)^2 \ dx$
$\ \ = int \ 4 sin^2 2x \ dx$

Next we use the identity:

$cos^2x -= cos^2x-sin^2x => sin^2x -= 1/2(1-cos2x)$

So we can write:

$I = int \ 4 sin^2 2x \ dx$
$\ \ = int \ 4 (1/2(1-cos4x)) \ dx$
$\ \ = int \ 2-2cos4x \ dx$

Which we can readily integrate:

$I = 2x - (2sin4x)/4 + C$
$\ \ = 2x - 1/2sin4x + C$

What is int 16sin^2 xcos^2 x dx ∫16sin2xcos2xdxint 16sin^2 xcos^2 x dx ?